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d. Two point charges, q1 = +25 nC and q2 = -75 nC, are separated by a distance of 3.0 cm. Find the magnitude and direction of; i. the electric force q1 exerts on q2 [5] ii. the force that q2 exerts on q1 [4] (take k = 9.0 x 109 N.m2 /C2 )

Sagot :

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Answer:

a) F₂₁ = 0.02 N, attracting.

b) F₁₂ = 0.02 N, attracting.

Explanation:

a)

  • The magnitude of the force that q₁ exerts on q₂ (F₂₁) is given by Coulomb's Law, as follows:

      [tex]F_{21} = k * \frac{q_{1} *q_{2}}{r_{12}^{2} } = 9e9 N.m2/C2 * \frac{(25e-9C)*(75e-9C)}{(0.03m)^{2}} = 0.02 N (1)[/tex]

  • Since q₁ and q₂ have opposite signs, the force between them will be always attractive, i.e., from q₂ towards q₁, along the line that joins both charges.

b)

  • The magnitude of the force on q₁ due to q₂ can be obtained applying Newton's 3rd Law, or using (1), because all parameters are the same, so F₁₂ (in  magnitude) = F₂₁ = 0.02 N
  • As we have already said, it must be opposite to the one found in a) so it must go from q₁ towards q₂, it is an attracting force also.